Use Green's theorem on ∮ (-y) dx + (x) dy around the rectangle 0 ≤ x… Step by Step

Step by step

The question

Question: use Green's theorem to find ∮ (−y) dx + (x) dy around the rectangle 0 ≤ x ≤ 2, 0 ≤ y ≤ 3.

∮ (−y) dx + x dy

around 0 ≤ x ≤ 2, 0 ≤ y ≤ 3

1 Find P and Q

The path goes once around the rectangle, counterclockwise.

The part before dx is P = −y.

P = −y

The one before dy is Q = x.

Q = x

2 Find the inside

Green's theorem: the line integral around equals the double integral inside of ∂Q/∂x minus ∂P/∂y.

∮ P dx + Q dy = ∬ (∂Q/∂x − ∂P/∂y) dA

Term by term.

x gives 1.

So ∂Q/∂x = 1.

∂Q∂x = 1

Term by term.

y gives 1.

So ∂P/∂y = −1.

∂P∂y = −1

Take away: 1 minus (−1).

∂Q∂x − ∂P∂y = 1 − (−1)

So the inside is 2.

∂Q∂x − ∂P∂y = 2

Careful!

The order matters.

Q by x, minus P by y.

The other way round changes the sign.

Careful: ∂Q/∂x first, then minus ∂P/∂y

3 Find the limits

Over the rectangle: y from 0 to 3 inside, x from 0 to 2 outside.

∮ = ∫ from 0 to 2 of ∫ from 0 to 3 of 2 dy dx

4 The inner integral

Integrate each term: the power goes up by 1, and we divide by the new power.

∫ from 0 to 3 of 2 dy = 2y from 0 to 3

Put in the top y = 3, then take away the bottom y = 0.

∫ from 0 to 3 of 2 dy = 2(3) − 2(0)

The bottom part is 0.

So it is 6.

∫ from 0 to 3 of 2 dy = 6

5 The outer integral

Integrate each term: the power goes up by 1, and we divide by the new power.

∫ from 0 to 2 of 6 dx = 6x from 0 to 2

Put in the top x = 2, then take away the bottom x = 0.

∫ from 0 to 2 of 6 dx = 6(2) − 6(0)

The bottom part is 0.

So it is 12.

∫ from 0 to 2 of 6 dx = 12

6 Check

Let's check by walking the four sides, one at a time.

Bottom 0, right 6, top 6, left 0.

On each side, only one of x and y moves.

bottom 0, right 6, top 6, left 0

Added up: 12.

The same as the double integral.

Right!

0 + 6 + 6 + 0 = 12

Let's look back.

First: find P and Q.

Then: find the inside.

Then: find the limits.

Then: the inner integral.

Then: the outer integral.

Last: check.

Answer